Question

A car of mass 10 kg moves on a circular track of radius 20m .if cofficient of friction is 0.3,then the maximum velocity with which the car can move is

Answer 1

$\huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}$

• To find : Maximum velocity with which car can move ?

• Given : mass of car = 10 kg

cofficeint of friction ,μ = 0.3

radius of circular track = 20 m

★ ${\red{\huge{\underline{\mathbb{Concept :-}}}}}$

The maximum velocity v with which the car can move is when the friction force will provide the necessary centripetal force.

${\purple {\huge{\underline{\mathbb{Answer:-}}}}}$

Frictional force

= μmg

$= 0.3 \times 10 \times 10$

$= 30 \: Newton$

Centripetal force

$= \frac{mv {}^{2} }{r}$

$= \frac{10 \times v {}^{2} }{20}$

$= \frac{v {}^{2} }{2}$

For maximum velocity

${\red{\boxed{\large{\bold{Frictional \: Force =\: Centripetal \: force }{}}}}}}$

$30 = \frac{v {}^{2} }{2}$

$v {}^{2} = 60$

$v = \sqrt{60}$

$v = 7.74 \: ms {}^{ - 1}$

Answer 2

first understand the concept......

the velocity of particle is totally dependent on centripetal force....more will be the force,,,more will be the velocity.....

but there is a limit of centripetal fotce,,if it become more then,the maximum frictional force then,then,the object will fall in toward the centre....and circular path will breakup...

so Maximum centripetal force=maximum frictional force

Now, let's move on the question

MV^2/R=.3×10×10

V^2=60

v=√60

v=7.76m/s

hope it helps you